Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.5 Factoring Sums or Differences of Cubes - 5.5 Exercise Set: 60

Answer

$-y(3x^2+3xy+y^2)$

Work Step by Step

The expressions $ x^3 $ and $ (x+y)^3 $ are both perfect cubes (the cube root is exact). Hence, $ x^3-(x+y)^3 $ is a $\text{ difference }$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x)^3-(x+y)^3 \\\\= [x-(x+y)][(x)^2+x(x+y)+(x+y)^2] \\\\= [x-x-y][x^2+x^2+xy+(x^2+2xy+y^2)] \\\\= -y(x^2+x^2+xy+x^2+2xy+y^2) \\\\= -y(3x^2+3xy+y^2) .\end{array}
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